Factors in the Dirichlet Function with Label 5,4 (1)

We are looking now at the Dirichlet series,

L(χ,s)

= 1 − 2-s − 3-s + 4-s + 6-s − 7-s − 8-s + 9-s + 11-s − 12-s − 13-s + 14-s + 16-s −  where χ is the Dirichlet character with label 5, 4.

So here when we divide each natural number by 5, when the remainder is either 2 or 3, we give it a negative sign; when the remainder is 1 or 4 we give it a positive sign and when the reminder is 0, we omit the number.

To calculate the zeros of this Dirichlet (5,4) function to t (on the imaginary scale) we calculate the corresponding zeros of the Riemann zeta function to 5t before then dividing the total number of zeros k by 5.

n	Rem. 2	Rem. 3	Total	Freq. of zeros	% acc.
50	45	34	79	72	91.1
100	102	83	185	172	93.0
150	161	135	296	282	95.3
200	228	191	419	399	95.2
250	296	246	542	521	96.1
300	367	309	676	648	95.9
350	439	370	809	777	96.0
400	513	435	948	910	96.0
450	588	500	1088	1044	96.0
500	663	565	1228	1182	96.3

In the above table, the 1^st column shows the accumulated number of factors (of a given type) up to n.

Then the 2^nd column shows the accumulated no. of factors that leave a remainder of 2 when divided by 5, while the 3^rd column shows the accumulated no. of factors that leave a remainder of 3 when divided by 5.

So for example 12 is made up of 5 divisors (i.e. factors) i.e. 2, 3, 4, 6 and 12. Because 1 is a factor of all numbers, we exclude it in this case.

So two of these i.e. 2 and 12 leave a remainder of 2 when divided by 5 and one i.e. 3 leaves a remainder of 3 when divided by 5. So 12 therefore contributes 2 factors to the Rem. 2 total and 1 factor to the Rem. 3 total.

The 4^th column then represents the total accumulation of factors (where the factors of a number leave a remainder of either 2 or 3 when divided by 5).

And remember that in the Dirichlet series (5,4) all natural numbers that leave a remainder of either 2 or 3 (when divided by 5) are given a negative sign.

The 5^th column then shows the frequency of zeros for this particular Dirichlet function with label (5, 4) up to a given no. which is calculated as follows.

Now if we look at numbers that leave a remainder of 2 or 3, when divided by 5, these constitute 2/5 (or 40%) of all natural numbers

Therefore to get an estimate of the accumulated factors of numbers up to n (that leave a remainder of 2 or 3 when divided by 5, we estimate the corresponding total accumulation of all factors up to 2n/5.

So for example   to estimate the accumulated factors of all numbers up to 100 (that leave a remainder of 2 or 3 on division by 5) we estimate the corresponding total of all factors up to 40.

There are 118 such factors up to (and including) 40.

The total of factors (that leave a remainder of 2 or 3 on division by 5) up to 100 is from Col. 3 of the table above 185.

Now in all these cases, where numbers carry positive or negative signs, the numbers associated with each sign will follow one of two distributions where the distribution will either be the standard Riemann distribution or alternatively the Dirichlet distribution associated with the specific function (5,4).

There are 118 such factors up to (and including) 40.

However the total of factors (that leave a remainder of 2 or 3 on division by 5) up to 100 is from Col. 3 of the table above 185.

So the discrepancy here is very marked.

Therefore we can safely assume that the numbers with negative signs i.e. that leave a remainder of 2 or 3 when divided by 5, are associated with the corresponding Dirichlet distribution (5, 4).

And as the conductor here is 5, to estimate the accumulated no. of factors (that leave a remainder of 2 or 3 on division by 5) to 100 we estimate the accumulated total of all factors to 200 i.e. 40 * 5) and then divide this total by 5.

So the total number of factors to 200 = 898, and 898/5 = 180 (to nearest integer)

And this compares very favourably with the actual no. of such factors to 100 = 185.

However we can use the frequency of Riemann zeta zeros directly to make a surprisingly accurate estimate of such factors.

In general terms to estimate the no. of all factors to n, we estimate the corresponding frequency of all zeros to t (where n = t/2π).

However in this case to estimate the frequency of Dirichlet zeros (5,4) to t, we estimate the frequency of Riemann zeros to 5t and then divide total by 5.

So if k = frequency of all zeros to 5t, then k/5 measures frequency of zeros to t

And once again to measure the frequency of factors that leave a remainder of 2 or 3 on division by 5 up to n, we measure the frequency of all factors to 2n/5.

And as n = t/2π, 2n/5 = t/5π.

Therefore to estimate for example accumulated total of factors (that leave a remainder of 2 or on division by 5) up to 100 we measure the Dirichlet zeros to 40 * 2π, which in turn means estimating the frequency of Riemann zeros (k) to 200 * 2π and then dividing by 5.

So the number of  Riemann zeros (k) to 200 * 2π  = 860 (to nearest integer).

And k/5 = no. of Dirichlet zeros to 40 * 2π  = 172.

And this is already a very good estimate (93% accurate) of the actual total of such factors i.e. 185.

Therefore because the frequency of zeros of each Dirichlet function is directly related to the corresponding frequency of zeros for the Riemann zeta function, we can always relate the accumulated total of factors for any Dirichlet function to the corresponding frequency of Riemann zeros, thereby using these latter well-known zeros as a goof estimate for the corresponding total of factors.

As we can see from the tables when we estimate the total no. of factors (leaving a remainder of 2 or on division by 5) up to 500, the corresponding estimate of Dirichlet (5,4) zeros to 2t/5 which in turn is related to the frequency of Riemann zeros (k) to 2t as k/5 already gives an estimate that is more than 96% accurate.

There is another fascinating relationship that is worth relating

In the table above, I have given separate totals factors that leave a remainder of 2 from those that leave a remainder of 3 (when divided by 5).

These two in turn follow a fascinating pattern.

As there are just two options here by which the natural numbers can be given a negative sign in the Dirichlet function (5,4) i.e. leaving a remainder of 2 or 3 on division by 5, this means that if we measure the total frequency of both types to n, then one the lower total will represent the frequency of factors to n/2 and the other the corresponding frequency from n/2 to n.

So in the above case the accumulated factors that leave a remainder of 3 (on division by 5) is lower than the corresponding accumulation of factors that leave a reminder of 3

Therefore the accumulated amount for the former with a remainder of 3, gives a good estimate for the total number of factors (with a remainder of either 2 or 3) up to n/2, with the accumulated amount with a remainder of 2 giving a good estimate of the total from /2 to n.

For example the total no. of factors to 500 (with a remainder of 2 or 3 on division by 5) = 1228.

Now the accumulated no. of factors (with a remainder of 3) to 500 = 565. And this corresponds well to the total no. of factors (with a remainder of either 2 or 3 to 250) = 542.

And the accumulated no. of factors (with a remainder of 2) to 500 = 663.

The total no. of factors from 250 to 500 = 1228 − 542 = 686. Again, this compares well with the reminder 2 figure, i.e. 663 (over 96% accurate).